Problem 103 - Chemistry: Principles and Reactions

Question

What is the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) ? You may assume that the first ionization is complete. The second ionization constant is \(0.010\).

Step-by-Step Solution

Answer

Answer: The pH of the 0.200 M H2SO4 solution is approximately 0.68.

Step 1: Determine the concentration of H+ from the first ionization of H2SO4

Since we're given that the first ionization is complete, we can directly calculate the concentration of H+ ions produced in this step. For every one mole of H2SO4, one mole of H+ ions and one mole of HSO4- ions are produced. Therefore, the concentration of H+ ions from the first ionization is equal to the concentration of H2SO4: 0.200 M.

Step 2: Determine the concentration of H+ from the second ionization of HSO4-

Now, we need to determine the concentration of H+ ions produced from the second ionization of HSO4-. The second ionization constant (K2) is given as 0.010. We can represent the second ionization as: HSO4- (aq) <=> H+ (aq) + SO42- (aq) Using the assumption that the initial concentration of H+ ions is entirely from the first ionization (as calculated in Step 1), we can set up the reaction quotient (Q) for the second ionization: Q = [H+][SO42-] / [HSO4-] Substituting K2 for Q and the initial concentrations into the equation, we get: 0.010 = ([0.200 + x] * x) / (0.200 - x) Where x represents the increase in [H+] and the decrease in [HSO4-] (i.e., the amount of HSO4- ionized in the second step).

Step 3: Solve for x

To solve for x, we can first assume that x is small compared to 0.200 (since K2 is small). This simplifies our equation to: 0.010 = (0.200 * x) / 0.200 Solving for x, we get: x = 0.010 M This value represents the additional concentration of H+ ions formed from the second ionization of HSO4-.

Step 4: Calculate the total concentration of H+ ions

Now, we can find the total concentration of H+ ions in the solution by adding the concentrations from both ionizations: [H+] = [H+ from first ionization] + [H+ from second ionization] [H+] = 0.200 M + 0.010 M [H+] = 0.210 M

Step 5: Calculate the pH of the solution

Finally, we can calculate the pH of the solution using the formula: pH = -log10([H+]) pH = -log10(0.210) pH ≈ 0.68 So the pH of the 0.200 M H2SO4 solution is approximately 0.68.

Key Concepts

Ionization of Sulfuric Acid

Sulfuric acid (H_2SO_4) is known for being a strong acid, meaning it dissociates readily in water. The ionization process involves step-by-step dissociation, releasing hydrogen ions (H^+). Sulfuric acid undergoes two distinct ionizations:

• The first ionization is complete in water. When sulfuric acid dissolves, it separates into H^+ ions and hydrogen sulfate ions (HSO_4^−):
  \[ \text{H}_2\text{SO}_4 \rightarrow \text{H}^+ + \text{HSO}_4^- \]
• The second ionization is partial. Here, HSO_4^− ions can further dissociate into H^+ and sulfate ions (SO_4^{2-}):
  \[ \text{HSO}_4^- \leftrightharpoons \text{H}^+ + \text{SO}_4^{2-} \]The second ionization isn't complete due to a smaller equilibrium constant (0.010). This means HSO_4^− doesn't entirely transform into SO_4^{2-} and H^+.

Understanding these ionizations is crucial for grasping how acidic the solution becomes and accurately determining the resulting hydrogen ion concentration.

Acid-Base Equilibrium

In chemistry, equilibrium describes the state of a reaction when both reactants and products exist in a stable balance. Although dynamically ongoing, the forward and backward reaction rates are equal. When talking about sulfuric acid, it is important to understand its place in the acid-base equilibrium.
The first ionization of sulfuric acid pushes more H^+ ions out due to complete dissociation. It shifts balance almost entirely to the products. However, the second ionization needs special attention, as it addresses the smaller equilibrium constant:

• The second equation,\( \text{HSO}_4^- \leftrightharpoons \text{H}^+ + \text{SO}_4^{2-} \)avails only a fraction of HSO_4^− dissociating.
• The equilibrium constant, K_2 = 0.010, suggests sparse promotion of ionization in the forward direction.

The primary focus is determining additional H^+ from the second stage, helping to finalize the total effective hydrogen ion concentration. This step-wise process evidences equilibrium's centrality in acid-base chemistry.

pH Scale

The is a convenient tool for gauging how acidic or basic a solution is, ranging from 0 to 14. It reflects the hydrogen ion concentration and helps classify the nature of the solution:

• A low pH (below 7) signals an acidic solution with high H^+ concentration. In our problem, sulfuric acid's pH can be expected to be particularly low, reflecting its strong acidic nature.
• A neutral pH at 7 means balanced hydrogen and hydroxide ion concentrations present as in pure water.
• High pH values (above 7) denote a basic solution with elevated OH^− concentration.

To compute the pH from the hydrogen ion concentration, use the formula:\[\text{pH} = -\log_{10}([\text{H}^+])\]Calculated pH, in this case, turns out to be remarkably low at approximately 0.68, owing to the significant H^+ concentration from the completed first ionization and substantial second ionization. This measure attests to sulfuric acid’s potency in acidic strength, vividly demonstrated through the pH scale.